# The Unapologetic Mathematician

## Mathematics for the interested outsider

Now that we know how to transform adjoints, we can talk about whole families of adjoints parametrized by some other category. That is, for each object $P$ of the parametrizing category $\mathcal{P}$ we’ll have an adjunction $F_P\dashv G_P:\mathcal{C}\rightarrow\mathcal{D}$, and for each morphism of $\mathcal{P}$ we’ll have a transformation of the adjunctions.

Let’s actually approach this from a slightly different angle. Say we have a functor $F:\mathcal{C}\times\mathcal{P}\rightarrow\mathcal{D}$, and that for each $P\in\mathcal{P}$ the functor $F(\underline{\hphantom{X}},P):\mathcal{C}\rightarrow\mathcal{D}$ has a right adjoint $G(P,\underline{\hphantom{X}})$. Then I claim that there is a unique way to make $G$ into a functor from $\mathcal{P}^\mathrm{op}\times\mathcal{D}$ to $\mathcal{C}$ so that the bijection $\phi_{C,P,D}:\hom_\mathcal{D}(F(C,P),D)\rightarrow\hom_\mathcal{C}(C,G(P,D))$ is natural in all three variables. Note that $G$ must be contravariant in $P$ here to make the composite functors have the same variance in $P$.

If we hold $P$ fixed, the bijection is already natural in $C$ and $D$. Let’s hold $C$ and $D$ fixed and see how to make it natural in $P$. The components $\phi_P:\hom_\mathcal{D}(F(C,P),D)\rightarrow\hom_\mathcal{C}(C,G(P,D))$ are already given in the setup, so we can’t change them. What we need are functions $\hom_\mathcal{D}(F(1_C,p),1_D):\hom_\mathcal{D}(F(C,P'),D)\rightarrow\hom_\mathcal{D}(F(C,P),D)$ and $\hom_\mathcal{C}(1_C,G(p,1_D)):\hom_\mathcal{C}(C,G(P',D))\rightarrow\hom_\mathcal{C}(C,G(P,D))$ for each arrow $p:P\rightarrow P'$.

For naturality to hold, we need $\hom_\mathcal{C}(1_C,G(p,1_D))\circ\phi_{P'}=\phi_P\circ\hom_\mathcal{D}(F(1_C,p),1_D)$. But from what we saw last time this just means that the pair of natural transformations $(F(1_C,p),G(p,1_D))$ forms a conjugate pair from $F(\underline{\hphantom{X}},P)\dashv G(P,\underline{\hphantom{X}})$ to $F(\underline{\hphantom{X}},P')\dashv G(P',\underline{\hphantom{X}})$. And this lets us define $G(p,1_D)$ uniquely in terms of $F(1_C,p)$, the counit $\epsilon'$ of $F(\underline{\hphantom{X}},P')\dashv G(P',\underline{\hphantom{X}})$, and the unit $\eta$ of $F(\underline{\hphantom{X}},P)\dashv G(P,\underline{\hphantom{X}})$ by using the first of the four listed equalities.

From here, it’s straightforward to show that this definition of how $G$ acts on morphisms of $\mathcal{P}$ makes it functorial in both variables, proving the claim. We can also flip back to the original viewpoint to define an adjunction between categories $\mathcal{C}$ and $\mathcal{D}$ parametrized by the category $\mathcal{P}$ as a functor from $\mathcal{P}$ to the category $\mathbf{Adj}(\mathcal{C},\mathcal{D})$ of adjunctions between those two categories.

July 31, 2007 - Posted by | Category theory

1. With every new post you make I’m finding out that I know almost nothing about adjoints!

Do you plan on talking about fibered categories at some point?

2. Maybe eventually I will. I’ve got a short-term goal for the category theory, though, before I can finally move on to topology.

And analysis.

And linear algebra.

And representation theory…

Comment by John Armstrong | July 31, 2007 | Reply

3. Suppose C is a monoidal category and F is the monoidal operation. Suppose further that the functor F(_,P) has a right adjoint, g(P) (contravariant) for each object P in C.

Now it seems that you are suggesting that there is a functor

G: F(C,C)–>C such that g(P) = G(P,_)

Assuming that I am understanding you correctly,

1) Can G be thought of as a right adjoint to F?
2) Is G(P,Q) necessarily an (the) exponential object Q^P? If it is, then C is, by definition, a closed category, right?
3) Can one start with a category C, a bifunctor G and define another functor F analogously? Presumably, G would have to satisfy properties similar to the monoidal operation but not quite dual properties.

dan

Comment by Dan | September 25, 2007 | Reply

4. Dan, I think your notation is a bit confused, but I can answer question 2 definitely. Yes, this is the definition of the exponential object, and in fact that’s why I gave this definition. If $\underline{\hphantom{X}}\otimes P$ has a right adjoint $E_P(\underline{\hphantom{X}}$ for all $P$, then these right adjoints fit together to make a functor $E:\mathcal{C}^\mathrm{op}\times\mathcal{C}\rightarrow\mathcal{C}$ so that $E(P,\underline{\hphantom{X}})=E_P(\underline{\hphantom{X}})$.

In this case, the functor isn’t the “right adjoint”, but the two functors together define a parametrized family of adjunctions.

And yes, you can start with something like the exponential and extract a monoidal structure given the correct left adjoints. In fact, I did just this just the other day when I talked about ordered linear spaces.

Of course, neither of these families of adjoints has to exist. But when they do they give us a lot of useful structure to work with.

Comment by John Armstrong | September 25, 2007 | Reply